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find the no. of solution of -x^2+x-1 = x^2008 + |x|

find the no. of solution of
-x^2+x-1 = x^2008 + |x|

Grade:12

2 Answers

Pratham Ashish
17 Points
11 years ago

hi rahul,

dis ques can be done by d approach of graph,,,

for LHS,

f = -x^2+x-1       ,f < 0 for all x <0,n f<0 for x>1, f >0 for 0<x>1........

 

since dy/dx < 0, n dis is a monotonus fxn so it decreases from +infinite to o at x=1,,,,,,,,,,

for rhs, f = x^2008 + |x|,   dis is a increasing fxn,,,,,n for all x f>0....

at x=1 , f=2,,at x =0,f=0,,,

so d two fxn can only intersect at one point in b/w 0<x>1...

so no. of soln = 1.

Rajesh Juluru
13 Points
10 years ago

for x>0,

-x^2-1=x^2008

x^2008+x^2+1=0

So no positive roots (since all are positive terms)

for x<0,

-x^2+2x-1=x^2008

x^2008+(x-1)^2=0

so no negative roots (since all are positive terms)

So i think number of solutions is zero

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